Electrons in Atoms: Introductory Comments by ROY66xy

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									         Electrons in Atoms: Introductory Comments
This unit is about where the electrons are located in an atom and some aspects of how
they behave. In a high school class, what is usually given are the answers and only a tiny
bit of why the answers are the way they are.

For example, you will soon learn the ground state electron configuration for sodium is 1s2
2s22p6 3s1. Once you learn the pattern, such answers will come easy. However, the
process by which the above answer was discovered was decades-long and involved
many, many people and lots of blind alleys and wrong turns. You simply won't have the
time to learn much about this process.

However, you will learn a little of it. That is why units like this one typically start out
with a discussion of light and what a spectrum is. The reason for this is that the above
configuration (and many others) was discovered through intensive study of the spectra of
the elements, principally hydrogen's spectrum.

All these discoveries about how electrons in an atom behave must be able to account for
the behavior of the elements. Why are some elements very chemically reactive? Why are
some not so reactive? Why does one member of an otherwise similar-behaving group of
elements engage in different behavior from the others? Why, where certain chemical
react, is one substance produced, but not another? Why cannot certain chemical
compounds be made? The list goes on and on and on, as you can well imagine.

I have no wise words that will help make this unit easy. It is not, because the ideas
involved are very complex. All I can say is, if you are stuck, do not give up. Take a
break, eat some ice cream and go back and try it again. You'll do just fine.

               A Brief (Incomplete) History of Light and Spectra

In 1913, Niels Bohr explained why the lines in the hydrogen spectrum are arranged the
way they are. His explanation places electrons in stable orbits, which he called
"stationary states" (violating well-known laws of science) and explains spectrum lines as
the energy difference resulting from the movement of an electron from a higher-energy
stationary state to a lower-energy one.

This moment not only ended almost 250 years of since the discovery of the spectrum, but
also was the beginning of fruitful ideas from one of the most profound intellects this
world has ever seen. (L. Brillouin tells that he had entered the room just as Sommerfeld
finished Bohr's article in the July 1913 issue of Philosophical Magazine. Sommerfeld
said "There is an extremely important paper here by N. Bohr which will be a milestone in
theoretical physics.") Bohr's work continues to have an impact on science today.

The best place to start, I suppose, is at the beginning.




                                              1
Matter and energy interact. Take a lump of metal and heat it. It glows, giving off light.
Burn some wood and one of the things it emittes is light. Place an object in the sunlight
and it gets hotter. Look at the sky after a rain and see a rainbow.

These things were known to the ancients. Today, we know that the energy given off by
matter creates a spectrum. This spectrum and the behavior of its colors are so important
to us that we have given its study a name -- spectroscopy.

The use of spectroscopy has mushroomed since its roots in the early 1800s. Today, nearly
200 years after its beginnings, research using spectroscopy and research into improving
the technique continues around the world.

What follows is in no way a complete history of spectral studies and the nature of light.
These are only some highlights I have selected.



1666

Isaac Newton allowed sunlight from a small, circular hole to fall on a prism, producing a
rainbow of color. Although the production of a rainbow by a clear crystal was known to
the ancients, it was Newton who showed that the colors did not originate in the crystal,
but rather were components of sunlight. This array of colors he called a spectrum.

Here is how the great man explained it:

       "And so the true Cause of the Length of that Image was detected to be no
       other, than that Light is not similar or Homogenial, but consists of Difform
       Rays, some of which are more Refrangible than others."

Newton's experimental arrangement is shown in the image below. It is from Voltaire's
Eléments de la Philosophie de Newton, published in 1738.




                                             2
In addition, Newton was able to produce a spectrum from white light and then recombine
the spectrum colors to get the white light back again.

Years later, in Opticks, Query 29, he gave the following mechanism for the fact that the
colors refracted from the greatest amount (violet) through blue, green, yellow, orange
(which he does not mention) and to the least amount (red):

       "Nothing is more requisite for producing all the variety of Colours, and
       degrees of Refrangibility than that the Rays of Light be Bodies of different
       Sizes, the least of which may take violet the weakest and darkest of the
       Colours, and be more easily diverted by refracting Surfaces from the right
       Course; and the rest as they are bigger and bigger, may make the stronger
       and more lucid colours, blue, green, yellow, and red, and be more and
       more difficultly diverted."

This explanation is a particle-based one: light particles are of different sizes.

1676

In Paris, Olaf (Ole) Roemer, working with Giovanni Cassini (the head of the
observatory), studies the variations in the calculated frequency of the planet Jupiter
eclisping its moon Io. To explain the variations, Cassini, in August 1675, writes:

       "The second irregularity in the motion of the first Jupiter's satellite may be
       due to [the fact] that light takes some time to reach us from the satellite,


                                               3
       and it takes from ten to eleven minutes to pass the distance equal to half
       the diameter of the Earth's orbit."

In September 1676, Roemer announced that the November 9 eclipse of Io by Jupiter
would occur 10 minutes later than that predicted. He was shown to be correct. In
December, he published a paper explaining his calculations, but is denounced by Cassini,
who becomes his opponent. Roemer also suffered because he was a Protestant and, in
France at that time, Protestants were just barely tolerated. In 1681, Roemer returned to
his native Denmark and was given many responsibilities in science and political
leadership. He is showered with many honors and awards and passed away September
19, 1710. Today he is known as the "Northern Archimedes" for his great range of
contributions in science.

In his paper on the speed of light, he never calculated the actual value. Since that time,
there have been speculative calculations based on Roemer's paper, but their values are
suspect. For example, one writer uses Romer's amount of time, but uses the modern
distance to the sun. Another writer assumed Roemer might have used some values that,
upon deeper analysis, he probably would not have. These two writers calculate that
Roemer's value for the speed of light (had he published one) would have been around
215,000 km/sec. The modern value is just under 300,000 km/sec.

In 1983, writing for the Journal for the History of Astronomy, Albert van Helden
speculated, using clues in Roemer's paper, that the calculated value - had there been one -
would have been around 135,000 km/sec.

1704-1730

Issac Newton publishes four editions of Opticks. Query 29 asks:

       "Are not the Rays of Light very small Bodies emitted from shining
       Substances?"

And Query 31 says:

       "Even the Rays of Light seem to be hard bodies; for otherwise they would
       not retain different properties in their different Sides."

In Opticks, he develops the particle theory of light (also called the corpuscular theory of
light). He is able to give plausible explanations for properties of light such as color,
reflection, and refraction.

He was not able to explain everything about light, diffraction bands outside the
geometrical shadow (discovered by Grimaldi in 1665) being one and Newton's rings
being another.




                                             4
An extremely important prediction implicit in Newton's particle theory is that, as light
moves from air to water, it SPEEDS up.

A wave theory of light existed in Newton's day. Its leading champion was Christian
Huygens, but the theory was incomplete. It only addressed a small fraction of the
phenomena Newton discussed and was difficult to understand. So, due to the wave
theory's poor explanatory power and Newton's great authority within the science world,
the particle theory of light reigned supreme.

1777

Carl W. Scheele (the discoverer of chlorine) investigated the effect of spectral colors on
the darkening of silver chloride. It was already known that silver chloride darkened upon
exposure to sunlight. Scheele found that, for the same exposure, the darkening was most
rapid at the violet end.

1800

William Herschel, the famous astronomer, studied the heating effects of the colors of the
spectrum. He did this by exposing thermometers to various colors of light. He found a
steady increase in heating power going towards the red. He also used a thermometer
placed just outside the red color as a control. To his amazement, the dark region just
outside the red color provided even more heating power than the red color.

In this manner, the infrared region was discovered.

1801

Johann Wilhelm Ritter verified Herschel's results and extended them. He placed silver
chloride outside the violet end of the spectrum and found even more darkening than
Scheele did 24 years earlier.

In this manner, the ultraviolet region of the spectrum was discovered.

This year also saw the discovery (by Thomas Young) of interference patterns caused by
light passing through a narrow slit. Up until this point, the wave theory of light
(Huygens) had had nothing convincing to offer in opposition to the particle theory of
light (Newton). Now it did, because you cannot explain interference using particles.

However, using waves, it was easy. Here is what Young said in the Bakerian Lecture,
November 29, 1801:

       "When two undulations, from different origins, coincide either perfectly or
       very nearly in direction, their joint effect is a combination of the motions
       belonging to both.




                                             5
       "Since every particle of the medium is affected by each undulation,
       wherever the directions coincide, the undulations can proceed no
       otherwise then by uniting their motions, so that the joint motion may be
       the sum or difference of the separate motions, accordingly as similar or
       dissimilar parts of the undulations are coincident."

However, despite the efforts of Young, the particle theory of light still remained the
accepted view of light's nature. In 1803, he writes:

       "The experiment of Grimaldi on the crested fringes within the shadow,
       together with several others of his observations equally important, has
       been left unnoticed by Newton. Those who are attached to the Newtonian
       theory of light, . . ., would do well to endeavor to imagine anything like an
       explanation of these experiments derived from their very own doctrines;
       and if they fail in the attempt, to refrain at least from idle declamation
       against a system which is founded on the accuracy of its application to all
       these facts, and to a thousand others of a similar nature."

To me, Young sounds a bit frustrated!!

1802

While verifying Ritter's results, William Wollaston also reported the existence of dark
lines in the spectrum of sunlight

He passed sunlight through a very narrow slit (no more than 0.05 inch) so that it fell on
the prism. (Up until this time, people had been using circular openings or relatively wide
slits. This method resulted in very impure spectra.)

Projecting the light over a distance of 10-12 feet, Wollaston saw red, yellowish-green,
green, blue and violet colors. He also reported seven dark lines. Five lines were reported
as being on the boundries between two colors, but two lines were within the color
boundries (specifically yellowish-green and blue).

1814-1823

Twelve years later, a young Joseph von Fraunhofer was looking for ways to check (and
improve) the quality of telescopes he was making. He rediscovered the dark lines in the
sun's spectrum while measuring the dispersive powers of various kinds of glass for light
of different colors. As he worked on this project, he noticed that a bright 'orange' line
(due to sodium, but he didn't know that) in the spectrum of the flame he was using was in
the same position as the dark D-line (see below). This same line had been observed in
flames from alcohol and sulfur as well as from candles.

Fraunhofer mapped out the 574 thin black lines that he observed in the sun's spectrum.
Eight of the most prominent lines were labeled A to G. Today, these lines are known as


                                             6
the Fraunhofer lines. The D-line was found to be two closely-spaced lines, subsequently
called D1 and D2. Its position turned out to be in the yellow portion of the spectrum, not
the orange that Fraunhofer wrote.

Here is what Fraunhofer drew:




In 1817 and 1823, Fraunhofer continued to report results from his research. He found:

       1) The spectrum of the moon showed the stronger lines of the sun in the
       same places.
       2) The spectra of Venus and Mars were faint, with some of the lettered
       lines being found.
       3) Several star spectra showed differences from the sun, although some of
       the same lines in the sun were identified in other stars.

In 1821, Fraunhofer reported on his first efforts to use a diffraction grating (rather than a
prism). Diffraction was discovered in 1665, but it received little attention until the early
1800s, when there was extensive discussion on the nature of light. The diffraction
grating, a series of closely spaced thin lines, was destined to play an important role in
future discoveries. Using his early diffraction gratings, Fraunhofer was able to measure
the wavelengths of the two sodium lines, obtaining values very close to the modern ones.

However, he could not explain why the dark lines where there. In fact, although spectral
research occupied him for much of his life, he never did find out how the lines that bear
his name were made.

The modern term for what he saw is an "absorption spectrum."

1815-1819

Augustin Fresnel independently rediscovered interference and begins to study (and
extend mathematically) the wave theory of light. In 1817, the French Academy of
Sciences decided to offer a prize for the best essay covering the wave theory of light. In
1819, Fresnel (one of two entries) wins the prize with a stunning 135-page
comprehensive treatment of the wave theory of light, refuting completely the particle
theory of light.


                                              7
On the judging panel was a mathematician of the first rank: Siméon Denis Poisson. He
also happened to be a very strong believer in Newton's particle theory of light and was
able, using Fresnel's mathematics, to derive a prediction he was sure would destroy the
wave theory of light: the famous Poisson's Spot. Here is what he said:

       "Let parallel light impinge on an opaque disk, the surroundings being
       perfectly transparent. The disk casts a shadow -- of course -- but the very
       centre of that shadow will be bright. Succinctly, there is no darkness
       anywhere along the central perpendicular behind an opaque disk (except
       immediately behind the disk). Indeed, the intensity grows continuously
       from zero right behind the thin disk. At a distance behind the disk equal to
       the disk's diameter, the intensity is already 80 per cent of what the
       intensity would be if the disk were absent. Thereafter, the intensity grows
       more slowly, approaching 100 per cent of what it would be if the disk
       were not present."

The judging committee chairman, François Arago arranged for an experiment to see if the
predicted spot was there. In his report, he wrote:

       ". . . Poisson has deduced from the integrals reported by the author the
       singular result that the centre of the shadow of an opaque circular screen
       must, when the rays penetrate there at incidences, which are only a little
       oblique, be just as illuminated as if the screen did not exist. The
       consequence has been submitted to the test of a direct experiment, and
       observation has perfectly confirmed the calculation."

So, with a prediction intended to destroy the wave theory, Poisson has succeeded in
advancing it greatly.

1826-1849

The next step in spectral analysis is due to John Herschel (son of William) and W.H. Fox
Talbot. They demonstrated, when a substance is heated and its light passed through a
spectroscope, that each element gave off its own set of characteristic bright lines of color.
In 1826, they wrote, "a glance at the prismatic spectrum of a flame may show it to
contain substances which it would otherwise require a laborious chemical analysis to
detect." The modern term for a spectrum of bright colored lines is "emission spectrum."

In 1832, David Brewster suggested that the dark lines in the solar spectrum might be
created by selective absorption of the light given off by the sun in its atmosphere. The
obvious question then became which chemical substance emitted which line or lines.

In 1833, William Miller demonstrated, when sunlight is passed through gases in the
laboratory, that additional dark lines appeared in the sun's spectrum. It was suggested that
the dark lines are due to the presence of gases on the sun. Miller also was one of the first
to take photographs of spectra.


                                              8
In 1834, Fox Talbot studied lithium and strontium, both of which give a red flame when
burned. He then wrote, "The prism betrays between them the most marked distinction
which can be imagined."

In 1840, John Herschel discovered that the Fraunhofer lines extended into the infrared,
the spectral region his father had discovered 40 years prior.

In 1842, A. Edmond Becquerel, photographed the solar spectrum plus its extension into
the ultraviolet.

In 1849, Jean Foucault, investigating the spectrum of an arc between two carbon
electrodes, noticed a line similar to the D line of the solar spectrum. He attempted to
superimpose the two spectra by passing the sun's rays through the arc and then through
the prism. This demonstrated that the lines were in the same place, since they
superimposed on each other.

He noticed something odd about the behavior of the D line. When the sunlight was shut
off, the D line appeared as a bright line in the arc spectrum. However, he wrote this: ". . .
the line D appears darker than usual in the solar light . . . Thus the arc presents us with a
medium which emits the D rays on its own account, and which at the same time absorbs
them when they come from another quarter."

Devoting his time to other work, he fails to follow up on this observation. It is left to
Gustav Kirchhoff, ten years later, to make the definitive discovery on the relationship
between the emission (bright-line) spectrum and the absorption (dark-line) spectrum

1849-1850

Up until this time, the only measurements of the speed of light were astronomically-
based, yielding only the speed of light in a vacuum. There was no way to test the
implications of Newton's 150-year-old particle theory of light, since it required
comparing the speed of light going from air to water. Lóon Foucault and Hippolyte
Fizeau, both in France, were working independently on measuring the speed of light and
thus be able to shed some light on this problem. Fizeau is the first (in 1849) to measure
the speed of light on the Earth's surface, but Foucault (in 1850) is the first to be able to
compare the two values needed to test Newton's particle theory for light.

The data is unequivocal: light travels SLOWER in water than it does in air. Newton's
particle theory cannot be true, as Fresnel so well demonstrated in 1819.

Foucault and Fizeau (younger by only 4 days) worked together in the early 1840's, but
parted ways without anger. Their competition was a model of scientific honesty and
integrity. In 1862, Foucault determined the speed of light to be 298,000 ± 500 km/sec.

     1859 - 1862: Gustav Kirchhoff (a physicist) and Robert Bunsen (a chemist)




                                              9
The relationship between emission & absorption spectra (1859).

Kirchhoff passed sunlight of moderate intensity through a flame containing lithium
chloride and observed the following:

       "One sees at the specific position [of the lithium line] a bright line on a
       dark background; for a greater intensity of incident sunlight, however,
       there appears at the same place a dark line, having exactly the same
       character as Fraunhofer's lines."

Kirchhoff also used incandescent lime, which was known to give off a continuous
spectrum. He passed the light from the lime through a sodium flame and then through a
prism. In the exact position of the D line of the solar spectrum, there appeared a dark line.

In this manner, he was able to demonstrate that hot gases absorbed the same wavelengths
of light they emitted. He knew then that sodium vapor must be present in the atmosphere
of the sun, absorbing the D line from the white light coming from the incandescent
surface of the sun.

He wrote:

       ". . . The dark lines of the solar spectrum, which are not caused by the
       earth's atmosphere, originate from the presence of those substances in the
       glowing solar atmosphere, which cause bright lines at the same place in
       the spectrum of a flame."

However, with the lithium flame, a new line appeared between the B and C lines
(remember these were labeled by Fraunhofer) that could not be associated with any of the
known dark lines. Hence, no lithium was present in the sun's atmosphere.

Chemical Analysis by Producing Spectra

Kirchhoff and Bunsen made a systematic survey involving many substances:

       "We have compared the spectra produced by the above-mentioned
       chlorine compounds with those obtained when the bromides, iodides,
       oxides, sulphates, and carbonates of the metals are brought into the flames
       of sulphur, carbon dioxide, aqueous alcohol, illuminating gas, carbon
       monoxide, hydrogen, and detonating gas.

       In this time-consuming, extensive research, which need not be presented
       here in detail, it came out that the variety of the compounds in which the
       metals were used, the differences in the chemical processes of the flames,
       and the great difference between their temperatures had no influence on


                                             10
       the position of the spectral lines corresponding to the individual metals."
       [emphasis theirs]

That is, every metal, no matter what compound it was in, gave the same spectrum.

Kirchhoff and Bunsen, in late 1860 discovered the element cesium by spectral analysis
and in early 1861 they discovered rubidium in the same matter. Both are named for the
color of the most prominent line in the spectrum (Latin rubidus, deepest red; caesius, sky
blue)



                          Brief Commentary before moving on

Please do not get the impression that the above story is the complete story of spectral
research or of the nature of light. I am only telling pieces of it as I build to Bohr's
explanation of the hydrogen spectra in 1913. For example, I am totally skipping over the
discovery of band spectra. This is an extremely important type of spectra, but it does not
fit into the story I am telling.

Line spectra, which I have been discussing above, were well-known to the scientific
community of the 1860s. About the same time as K&B, a different type of spectra (today
called 'band' spectra) was discovered. Of course, discussion arose as to how the two were
different. By 1875, the correct answer was in hand. It simply was that line spectra are
produced by free elements (atoms) and band spectra by compounds (molecules).

However, band spectra do not play a role in the direction I am going, so this is the last
you'll hear of them from me.

          A Brief (Incomplete) History of Light and Spectra: Part 2

1862 Anders Jonas Ångström identified three lines in the visible portion of the hydrogen
emission spectrum, a red line, a blue-green line and a violet line. Later on, he identifies
the violet line as being two lines very close together.

1868 Ångström introduced a great improvement. In his book Recherches sur le spectre
solaire, which contains an atlas of the solar spectrum with the wavelengths of over a
thousand lines obtained by the use of diffraction gratings, he expressed his wavelengths
in units of 10¯10 m. Before this time scales with arbitrary units had been used. This
Ångström unit is still widely used.

1871

Ångström measured the wavelengths on the four visible lines of the hydrogen spectrum.
In modern values, they are:



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                                             Wavelength
                                Line color
                                           (in Ångströms)
                                red        6562.852 Å
                                blue-green 4861.33 Å
                                violet     4340.47 Å
                                violet     4101.74 Å

1873

James Clerk Maxwell, a Scottish physicist, succeeded in unifying electricity and
magnetism. He did this by showing that the two forces have a common origin. Since that
time, we speak of electromagnetism. In his theory, it was clear that the emission of
radiation by matter must be as a result of the acceleration of electric charges. It seemed
that within matter there were mobile electrical charges that, as they moved in some
fashion, produced the spectrum.

However, here is where his theory failed. It predicted that the spectrum could be resolved
into sets of spectral lines, each consisting of a fundamental frequency (let's call it nνo)
and its harmonics. The fundamental would be when the n = 1 and the harmonics would
be where n = 2, , 4, . . . . In no case did the spectral lines correspond to what the theory
predicted. The (predicted, but still undiscovered in 1873) mobile charges DID NOT obey
the rules that Maxwell had discovered.

[I hope it is clear to my readers that these mobile charges are now called electrons and
that there is a set of rules the electrons obey. Bohr, in 1913, will begin the process of
discovering exactly what those rules were.]

1880s to 1908

During this time, many empirical regularities were found in spectral lines. What this
means is that some spectral lines could be related to others by an equation, but no one
knew why the equation worked. The most famous of this time was the Balmer formula in
1885.

Exactly what is the Balmer formula? It gave a simple, mathematical relationship between
the wavelengths of the four lines of the visible spectrum of hydrogen, measured by
Ångström in 1871. The equation shows that the four numbers are related in some way,
but in 1885 no one knew what that relationship was.

1896

Pieter Zeemen discovered the effect named for him: the Zeeman Effect. He discovered
that spectral lines split into very closely spaced groups when the light was passed through
a magnetic field first.



                                             12
For example, the red line of the hydrogen spectrum was known to be 6562.852 Å, but
that was with no magnetic filed present between the hydrogen source and the measuring
device (called a spectroscope). However, pass the light through a magnetic field on its
way from source to instrument, that's a different story. That single red line split (I think
into three, but I'm not sure) into several lines that were very closely spaced around the
6562.852 Å value.

Zeeman, of course, published the results of his work and others around the world began
to look at other lines in the spectrum of hydrogen and other substances. It turns out that
almost every line split in some fashion. Some split into three lines, others into five. Some
split into two and others into two groups of two. And the individual lines in the resulting
group did not all have the same intensity of color. Also, if the magnetic field was rotated
90 degrees, sometimes different splitting resulted.

All in all, it was very, very difficult to figure out what was going on. However, a correct
explanation was eventually arrived at, although it took a number of years.

A point of interest: Zeeman (a native of Holland) received a promotion to a more well-
known university as a result of this discovery. (He also won the Nobel Prize.) However,
his new university had older and poorer equipment than his old job and he could not
reproduce his experiment until better equipment was purchased.

1913

Niels Bohr publishes an explanation of the hydrogen spectrum, which today we call "The
Bohr Model of the Atom." In it, he uses the standard concepts of the time, but with a
twist. He uses the "quantum hypothesis," introduced in 1900 by Max Planck and makes a
critical assumption about how the lines in the spectrum are generated.



I will end my (incomplete) spectrum history here. There is more, much more, but that is a
story for another day. However, just a taste:

       (1) It turned out that there were splittings within the line splittings
       discovered by Zeeman. This resulted in the discovery of "spin" in 1925.
       (2) It turned out that the lines were not exactly in the places predicted, but
       were offset by a tiny, tiny amount. This was demonstrated by Willis
       Lamb, who received a Nobel Prize for discovering the "Lamb Shift." The
       Lamb Shift is due to the electrons interacting with the nucleus and his
       1949 experiment is considered a classic.

 Two Equations Governing Light's Behavior: Part One
                      λν = c


                                             13
There are two equations concerning light that are usually taught in high school.
Typically, both are taught without any derivation as to why they are the way they are.
That is what I will do in the following.

Equation Number One: λν = c

Brief historical note: I am not sure who wrote this equation (or its equivalent) first. The
wave theory of light has its origins in the late 1600's and was developed mathematically
starting in the early 1800's. It was James Clerk Maxwell, in the 1860's, who first
predicted that light was an electromagnetic wave and computed (rather than measured) its
speed. By the way, the proof that light's speed was finite was published in 1676 and the
first reliable measurements of the speed of light, ones that were very close to the modern
value, took place in the late 1850's.

Each symbol in the equation is discussed below. Also, right before the example problems,
there is a mention of the two main types of problems teachers will ask using the equation.
I encourage you to take a close look at that section.

1) λ is the Greek letter lambda and it stands for the wavelength of light. Wavelength is
defined as the distance between two successive crests of a wave. When studying light, the
most common units used for wavelength are: meter, centimeter, nanometer, and
Ångström. Even though the official unit used by SI is the meter, you will see
explanations and problems, which use the other three. Less often, you will see other units
used; picometer is the most common one among the less-often used wavelength units.
Ångström is a non-SI unit commonly included in discussions of SI units because of its
wide usage.

Keep in mind these definitions:

       one centimeter equals 10¯2 meter
       one nanometer equals 10¯9 meter
       one Ångström equals 10¯8 centimeter

The symbol for the Ångström is Å.

Most certainly, you will need to move easily from one unit to the other. For example,
notice that 1 Å = 10¯10 meter. This means that 10 Å = 1 nm. So, if you are given an
Ångström value for wavelength and a nanometer value is required, divide the Ångström
value by 10. If you can't make easy transitions between various metric units, you'd better
go back to study and practice that area some more.

2) ν is the Greek letter nu. It is NOT the letter v, it is the Greek letter nu!!! It stands for
the frequency of the light wave. Frequency is defined as the number of wave cycles
passing a fixed reference point in one second. When studying light, the unit for frequency
is called the Hertz (its symbol is Hz). One Hertz is when one complete cycle passes the
fixed point, so a million Hz is when the millionth cycle passes the fixed point.


                                              14
There is an important point to make about the unit on Hz. It is NOT commonly written as
cycles per second (or cycles/sec), but only as sec¯1 (more correctly, it should be written
as s¯1; you need to know both ways). The "cycles" part is deleted, although you may see
an occasional problem, which uses it.

A brief mention of cycle: imagine a wave, frozen in time and space, where a wave crest is
exactly lined up with our fixed reference point. Now, allow the wave to move until the
following crest is exactly lined up with the reference point, and then freeze the wave in
place. This is one cycle of the wave and if all that took place in one second, then the
frequency of the wave is 1 Hz.

In any event, the only scientifically useful part of the unit is the denominator and so "per
second" (remember, usually as s¯1) is what is used. The numerator "cycles" is not needed
and so its presence is simply understood and, if writing a fraction is necessary, a one
would be used, as in 1/sec.

3) c is the symbol for the speed of light, the speed at which all-electromagnetic radiation
moves when in a perfect vacuum. (Light travels slower when passing through objects
such as water, but it never travels faster than when in a perfect vacuum.)

Both ways shown below are used to write the value. You need to be aware of both:

       3.00 x 108 m/s
       3.00 x 1010 cm/s

The actual value is just slightly less, but the above values are the one generally used in
introductory classes. (sometimes you'll see 2.9979 rather than 3.00.) Be careful about
using the exponent and unit combination. Meters are longer than centimeters, so there are
less of them used above.

Since there are two variables (λ and ν), we can have two types of calculations: (a) given
wavelength, calculate frequency and (b) given frequency, calculate wavelength. By the
way, these are the two types of problems teachers generally ask on the test.

Note that we can easily rearrange the main equation to fit these two types of problems:

       (a) given the wavelength, calculate the frequency; use this equation: ν = c /
       λ

       (b) given the frequency, calculate the wavelength; use this equation: λ = c
       /ν



Example problem #1: What is the frequency of red light having a wavelength of 7000
Å?


                                             15
The solution below depends on converting Å into cm. This means you must remember
that the conversion is 1 Å = 10¯8 cm. The solution:

                        (7000 x 10¯8 cm) (x) = 3.00 x 1010 cm/sec

Notice how I did not bother to convert 7000 x 10¯8 into scientific notation. If I had done
so, the value would have been 7.000 x 10¯5.

Note also that I effectively consider 7000 to be 4 significant figures. I choose to do this
because I know wavelength measurements are very accurate and that 6, 7, or even 8 sig
figs are possible. At an introductory level, you will not know this, so that is why I am
telling you here. Also, the value for the speed of light is known to nine significant
figures, as in 299,792,458 m s¯1. However, 3.00 is good enough for introductory work.

The answer is 4.29 x 1014 s¯1

Example problem #2: What is the frequency of violet light having a wavelength of 4000
Å?

The solution below depends on converting Å into cm. This means you must remember
that the conversion is 1 Å = 10¯8 cm. The solution:

                        (4000 x 10¯8 cm) (x) = 3.00 x 1010 cm/sec

Notice how I did not bother to convert 4000 x 10¯8 into scientific notation. If I had done
so, the value would have been 4.000 x 10¯5. Note also that I effectively consider 4000 to
be 4 significant figures.

The correct answer is 7.50 x 1014 s¯1

Be aware that the range of 4000 to 7000 Å is taken to be the range of visible light. Notice
how the frequencies stay within more-or-less the middle area of 1014, ranging from 4.29
to 7.50, but always being 1014. If you are faced with this calculation and you know the
wavelength is a visible one (say 555 nm), then you know the exponent on the frequency
MUST be 1014. If it isn't, then YOU (not the teacher) have made a mistake.

Example problem #3: What is the frequency of EMR having a wavelength of 555 nm?
(EMR is an abbreviation for electromagnetic radiation.)

First, let us convert nm into meters. Since one meter contains 109 nm, we have the
following conversion:

       555 nm x (1 m / 109 nm)

       555 x 10¯9 m = 5.55 x 10¯7 m




                                             16
Inserting into λν = c, gives:

       (5.55 x 10¯7 m) (x) = 3.00 x 108 m s¯1

       x = 5.40 x 1014 s¯1

Example problem #4: What is the wavelength (in nm) of EMR with a frequency of 4.95
x 1014 s¯1?

Substitute into λν = c, as follows:

       (x) (4.95 x 1014 s¯1) = 3.00 x 108 m s¯1

       x = 6.06 x 10¯7 m

Now, we convert meters to nanometers:

       6.06 x 10¯7 m x (109 nm / 1 m) = 606 nm

Example problem #5: What is the wavelength (in both cm and Å) of light with a
frequency of 6.75 x 1014 Hz?

The fact that cm is asked for in the problem allows us to use the cm/s value for the speed
of light:

       (x) (6.75 x 1014 s¯1) = 3.00 x 1010 cm s¯1

       x = 4.44 x 10¯5 cm

Next, we convert to Å:

       (4.44 x 10¯5 cm) x (1 Å / 10¯8 cm) = 4440 Å

                  Calculate the frequency when given the wavelength.

1) Calculate the frequency of EMR that has a wavelength of 1.315 micrometers.

2) Calculate the frequency of radiation with a wavelength of 442 nm.

3) Calculate the frequency of radiation with a wavelength of 4.92 cm.

4) Calculate the frequency of radiation with a wavelength of 8973 Å.

5) Calculate the frequency of radiation with a wavelength of 4.55 x 10¯9 cm.

          Solutions to 'Calculate the frequency when given the wavelength'


                                            17
General comments:

To solve these problems, you must convert the wavelength to meters before using λν = c.
The reason for meters is that I will almost always use 3.00 x 108 m s¯1 for the speed of
light.

A bit of a warning: there could be a problem worded in such a way for which 3.00 x 1010
cm s¯1 (for the speed of light) is the better-suited value. Some examples are below.
Sometimes a teacher might supply the centimeter value in the problem, but the meter
value would be better suited for the solution. Be careful on your units!

These problems sometimes require metric conversions. If you wish to review metric
conversions, click here.



1) EMR means electromagnetic radiation.

First, convert μm to m:

       1.315 μm x (1 m / 106 μm) = 1.315 x 10¯6 μm

Then, substitute into λν = c:

       (1.315 x 10¯6 μm) (x) = 3.00 x 108 m s¯1

       x = 2.28 x 1014 s¯1

2) First, convert nm to m:

       442 nm x (1 m / 109 nm) = 4.42 x 10¯7 m

Then, substitute into λν = c:

       (4.42 x 10¯7 m) (x) = 3.00 x 108 m s¯1

       x = 6.79 x 1014 s¯1

3) Since the wavelength is already in cm, we can use c = 3.00 x 1010 cm s¯1 and not have
to do any conversions at all.

       (4.92 cm) (x) = 3.00 x 1010 cm s¯1

       x = 6.10 x 109 s¯1




                                            18
By the way, this is a low frequency wave, commonly called a radio wave. With radio
waves, the unit MHz (mega hertz) is often used. The above value would be 6100 MHz.

4) Since 1 Å = 10¯8 cm, therefore 8973 Å = 8973 x 10¯8 cm. Converting to scientific
notation gives 8.973 x 10¯5 cm. This is another place where the cm s¯1 value for c can be
used, since Å converts to cm very easily.

       (8.973 x 10¯5 cm) (x) = 3.00 x 1010 cm s¯1

       x = 3.34 x 1014 s¯1

5) See comments in the solution to 3). The solution for 5) is left to the reader.

Calculate the wavelength when given the frequency.

1) Light with a frequency of 7.26 x 1014 Hz lies in the violet region of the visible
spectrum. What is the wavelength of this frequency of light? Answer in units of nm.

2) When an electron beam strikes a block of copper, x-rays of frequency 1.07 x 1019 Hz
are emitted. What is the wavelength of these x-rays? Answer in units of pm.

3) Calculate the wavelengths (in meters) of radiation a frequency of 5.00 x 1015 s¯1.

4) Calculate the wavelengths (in meters) of radiation a frequency of 2.11 x 1014 s¯1.

5) Calculate the wavelengths (in meters) of radiation a frequency of 5.44 x 1012 s¯1.

6) What is the wavelength of sound waves having a frequency of 256.0 sec¯1 at 20 °C?
Speed of sound = 340.0 m/sec. (The problem is about sound, but this does not change the
basic idea of the equation "wavelength times frequency = speed."

Solutions to 'Calculate the wavelength when given the frequency'

General comments:

These problems sometimes require metric conversions. If you wish to review metric
conversions, click here.



1) We first need to substitute into λν = c:

       (x) (7.26 x 1014 s¯1) = 3.00 x 108 m s¯1

       x = 4.13 x 10¯7 m



                                              19
Since the question asks for the answer in nm, we must convert from m to nm:

       4.13 x 10¯7 m x (109 nm / 1 m) = 413 nm

2) We first need to substitute into λν = c:

       (x) (1.07 x 1019 s¯1) = 3.00 x 108 m s¯1

       x = 2.80 x 10¯11 m

Since the question asks for the answer in pm, we must convert from m to pm:

       2.80 x 10¯11 m x (1012 pm / 1 m) = 28 pm

By the way, if question 2 had asked for the answer in nm, it would have been 0.0280 nm.
Notice that the wavelength unit in this question was deliberately picked to give a whole
number. Generally speaking, the wavelength unit is picked to give a whole number, be it
tens, hundreds or thousands.

3) to 5) These problems are solved by substituting the frequency into λν = c, just like in
1) and 2) above. Note that each problem asks for the wavelength in meters. Only the
answers follow:

       3) 1.50 x 10¯7 m

       4) 1.42 x 10¯6 m

       5) 5.51 x 10¯5 m

If 3) to 5) were asked for in nm, you would multiply the above meter values by 109. See
1) above for an example. The answer, to give one example, for 5) would be 5510 nm.

6) (x) (256.0 sec¯1) = 340.0 m/sec

The answer, to four sig figs, is 1.328 m. In centimeters, 132.8 and in Ångströms, 1.328 x
1010

 Two Equations Governing Light's Behavior: Part Two
                      E = hν
There are two equations concerning light that are usually taught in high school.
Typically, both are taught without any derivation as to why they are the way they are.
That is what I will do in the following.

Equation Number Two: E = hν


                                              20
Brief historical note: It is well-known who first wrote this equation and when it
happened. Max Planck is credited with the discovery of the "quantum," the discovery of
which took place in December 1900. It was he who first wrote the equation above in his
announcement of the discovery of the quantum.

1) E is the energy of the particular quantum of energy under study. When discussing
electromagnetic quanta (of which light is only one example, x-rays and radio waves being
two other examples), the word photon is used. A photon (the word is due to Albert
Einstein) is a quantum of electromagnetic energy. The word quantum (quanta is the
plural) is usually used in a more general sense, to describe various ideas of quantum
theory or even, as I just did, to describe the entire theory itself.

2) h stands for a fundamental constant of nature now known as Planck's Constant. By the
way, the discovery of the quantum had, and continues to have, many profound effects.
Enough so that all of science (especially physics) before 1900 is referred to as "classical"
and the science since 1900 is called "modern."

The value for Planck's Constant is 6.6260755 x 10¯34 Joule second. Please note that the
unit is Joule MULTIPLIED BY second. It is not a division, both Joule and second are in
the numerator.

3) ν is the frequency of the particular photon being studied. The discussion about
frequency above applies here.

Before going on, I want to discuss one little issue (heh, heh, heh). Frequency is a wave-
based idea. What is it doing in a particle-based idea like the quantum? Good question. So
much so that the term "wave packet" is often used in discussing these ideas. Indeed,
modern science now speaks of "wave-particle duality" rather than "Light is a wave" or
"Light is a particle." This whole area is profound and can lead to years of probing
discussion. Albert Einstein and Niels Bohr (who were great friends) discussed these
issues (and more) often over a period of many years, especially in the late 1920's and
early 1930's. Their discussions are still important enough to merit historical study today.
Every year, several books are published which delve into one or more of the implications
of "wave-particle duality."

Example problem #1: (a) Identify λν = c as either a direct or inverse mathematical
relationship. (b) Do the same for E = hν. (c) Write a mathematical equation for the
relationship between energy and wavelength. (d) Identify the equation for (c) as either
direct or inverse.

The answers:

       (a) λν = c is an inverse relationship. As one value (say the wavelength)
       goes up, the other value (the frequency) must go down. Why? Because the
       product of the two must always equal the same value, c, which is a
       constant.


                                             21
       (b) E = hν is a direct relationship. As the frequency increases, so does E.
       Why? Because h remains constant. Notice that in this equation the two
       quantities which can change (ν and E) are on DIFFERENT sides of the
       equation whereas in the inverse relationship in (a), both λ and ν are on the
       same side, with the constant on the opposite side.

       (c) Rearrange the light equation thusly:

                                          ν=c/λ

       Substitute for ν in E = hν:

                                         E = hc / λ

       Rearrange to group the variables and the constants:

                                          Eλ = hc

       (d) The mathematical relationship is an inverse one. Note the equation's
       similarity to λν = c, with two values that can vary on the left side and a
       constant (h times c) on the right.

Example problem #2: How many Joules of energy are contained in a photon with λ =
550 nm? How many kJ/mol of energy is this?

I will first do the problem step by step, then in a more combined way.

Use ν = c / λ to get the frequency:

                          x = (3.00 x 108 m s¯1) / (550 x 10¯9 m)

This equals 5.4508 x 1014 s¯1. I left a couple extra digits in the answer and notice that the
wavelength is not in scientific notation. Why? Think about how I got from nm to m for
the value used. (I also used 299,792,458 in the calculation, not 3.00 x 108. Sorry!! Using
3.00 x 108 gives 5.454 x 1014 s¯1.)

Now use E = hν to get the energy:

                      x = (6.6260755 x 10¯34 J s) (5.4508 x 1014 s¯1)

This equals 3.612 x 10¯19 J. Most emphatically, this is not the final answer. This is the
energy for one photon!!

The last step is to find the kilojoules for one mole and for this we use Avogadro's
Number:


                                             22
                        x = (3.612 x 10¯19 J) (6.022 x 1023 mol¯1)

Dividing the answer by 1000 to make the change to kilojoules, we get 217.5 kJ/mol. Yes,
this is my final answer!

A more condensed way would be to use Eλ = hc and solve for E, then multiply the
answer times Avogadro's Number.

Example problem #3: How many kJ/mol (of photons) of energy is contained in light
with a wavelength of 496.36 nm?

The answer is 241.00 kJ/mol.



Three more problems:

1) What is the energy of a photon of green light with a frequency of 5.76 x 1014 s¯1.

2) A particular x-ray has a wavelength of 1.2 Å. Calculate the energy of one mole of
photons with this wavelength.

3) When excited, some atoms produce an emission with a frequency of 7.25 x 1012 Hz.

       (a) calculate the energy, in Joules, for one photon with this frequency.
       (b) calculate the energy, in kJ/mol.
       (c) Is this light visible? Why or why not?

More problems which use E = hν

1) What is the energy of a photon of green light with a frequency of 5.76 x 1014 s¯1.

       Solution:

               x = (6.626 x 10¯34 J s) (5.76 x 1014 s¯1)

               x = 3.82 x 10¯19 J

       Comment: all frequencies of visible light will have an energy in the 10¯19
       J range of values. If you wish to, you may calculate this for yourself. The
       wavelength range of visible light is taken to be from 400 nm to 700 nm.
       This translates (more-or-less) to a range from 5 x 10¯19 J down to 3 x
       10¯19 J.

2) A particular x-ray has a wavelength of 1.2 Å. Calculate the energy of one mole of
photons with this wavelength.


                                            23
       Solution:

       1.2 Å x (10¯8 cm / 1 Å) = 1.2 x 10¯8 cm

       (x) (1.2 x 10¯8 cm) = (6.626 x 10¯34 J s) (3.00 x 1010 cm s¯1)

       Comment: I used Eλ = hc. Note also that I used 3.00 x 1010 cm s¯1 for the
       speed of light. I did this because the 1.2 Ångstrom value for the
       wavelength converts very easily into cm. There was no need to take the
       wavelength to meters.

       x = 1.66 x 10¯15 J

       This is the energy for one photon. To get energy per mole, multiply the
       above value by Avogadro's number:

       (1.66 x 10¯15 J) (6.022 x 1023 mol¯1) = 9.98 x 108 J mol¯1

       This value is usually reported in kJ per mole: 9.98 x 105 kJ mol¯1

3) When excited, some atoms produce an emission with a frequency of 7.25 x 1012 Hz.

       (a) calculate the energy, in Joules, for one photon with this frequency.
       (b) calculate the energy, in kJ/mol.
       (c) Is this light visible? Why or why not?

       Comments on the solution:

              (a) use E = hν

              (b) use Avogadro's number as well as the answer from (a).
              Make sure to convert from the J value (which is what you'll
              calculate) to the kJ value.

              (c) Calculate the wavelength using λν = c. What you need
              to do is compare the wavelength you calculate to the
              commonly accepted range of visible wavelengths, which is
              400 nm to 700 nm. The wavelength you calculate will
              probably be in meters, so you will need to convert it to nm,
              then compare.

4) The radioactive isotope Thallium-201 is used in medical diagnosis and treatment. A
gamma ray emitted by an atom of Thallium-201 has an energy of 0.167 million electron-
volts. (1 MeV is 1 x 106 eV and 1 eV = 1.6022 x 10¯19 J). What is the frequency in Hz of
this gamma ray?




                                            24
Solution

       The first part converts the MeV value into Joules:

       0.167 MeV = 0.167 x 106 eV = 1.67 x 105 eV

       1.67 x 105 eV times 1.6022 x 10¯19 J/eV = 2.675674 x 10¯13 J

       use E = hν:

       2.675674 x 10¯13 J = (6.6260755 x 10¯34 J s) (x)

       Solve for x

                         The Hydrogen Spectrum
Go back and read through A Brief History of Spectroscopy up to 1862 if you haven't
done so already.

An emission spectrum of an element has lines in it, which are against a black
background. If those lines are in the visible spectrum, then we see them as specific colors.
Outside the visible spectrum, the lines only appear as lines on a photograph.

The emission spectrum of hydrogen occupies a very important place in the history of
chemistry and physics. Niels Bohr, in 1913, will use the hydrogen spectrum to start on
the road to explaining how electrons are arranged in an atom.

Here is a drawing of the visible spectrum of hydrogen:




The left edge ends at a wavelength of 4000 Ångströms and the right edge ends at 7000
Ångströms.

(The Ångström is equal to 10¯8 cm. and its symbol is Å. It was established by Anders
Ångström and he named it after himself. It is not an official SI unit, but its use is
tolerated. The more modern wavelength unit is nanometers, symbol nm. 4000 Å = 400
nm, 7000 Å = 700 nm.)




                                            25
The four colored lines are the four visible lines in the hydrogen spectrum. Their
wavelenghts in Ångströms are shown in the diagram. There is one red line (6562.852 Å),
one blue-green line (4861.33 Å), and two violet lines (4340.47 Å and 4101.74 Å).

In 1862, Anders Ångström discovered three lines and later on found the 4th line (the
4101.74 violet line). By 1871, he measured all four wavelengths to a high degree of
accuracy.

This next diagram is a photograph of a more complete hydrogen spectrum:




Hα, Hβ, Hγ, and Hδ are the official designations for the 4 lines of the visible portion of the
spectrum. All the other lines to the left of these four are in the ultraviolet portion of the
spectrum and so would not be visible to the eye. However, they can be photographed, as
they were here.

This grouping of lines has a name. It is called the Balmer Series. In 1885, Johann Balmer
was able to calculate the four visible lines' wavelengths using one formula, now called
the Balmer Formula. Notice how the lines get closer and closer to each other. This will
become important later.

                                  The Balmer Formula

Leading up to the Formula: 1869 - 1882

In the years after the work of Kirchhoff and Bunsen, the major goal in spectroscopy was
to determine the quantitative relationships between the lines in the spectrum of a given
element as well as relationships between lines of different substances.

For example, George Johnstone Stoney in 1869 speculated that spectra arose from the
internal motions of molecules. However, his mathematical theory was rejected and in
1881, Arthur Schuster concluded:

       "Most probably some law hitherto undiscovered exists . . . . "

One year later, Schuster added:

       "It is the ambitious object of spectroscopy to study the vibrations of atoms
       and molecules in order to obtain what information we can about the nature



                                              26
       of forces which bind them together . . . But we must not too soon expect
       the discovery of any grand or very general law, for the constitution of
       what we call a molecule is no doubt a very complicated one, and the
       difficulty of the problem is so great that were it not for the primary
       importance of the result which we may finally hope to obtain, all but the
       most sanguine might well be discouraged to engage in an inquiry which,
       even after many years of work, may turn out to have been fruitless.

       . . . In the meantime, we must welcome with delight even the smallest step
       in the desired direction."

The Balmer Formula: 1885

On June 25, 1884, Johann Jacob Balmer took a fairly large step forward when he
delivered a lecture to the Naturforschende Gesellschaft in Basel. He first represented the
wavelengths of the four visible lines of the hydrogen spectrum in terms of a "basic
number" h:




Balmer recognized the numerators as the sequence 32, 42, 52, 62 and the denominators as
the sequence 32 - 22, 42 - 22, 52 - 22, 62 - 22.

So he wound up with a simple formula which expressed the known wavelengths () of
the hydrogen spectrum in terms of two integers m and n:




For hydrogen, n = 2. Now allow m to take on the values 3, 4, 5, . . . . Each calculation in
turn will yield a wavelength of the visible hydrogen spectrum. He predicted the existence
of a fifth line at 3969.65 x 10¯7 mm. He was soon informed that this line, as well as
additional lines, had already been discovered.

Here are some calculations using Balmer's formula.

At the time, Balmer was nearly 60 years old and taught mathematics and calligraphy at a
high school for girls as well as giving classes at the University of Basle. Balmer was very




                                            27
interested in mathematical and physical ratios and was probably thrilled he could express
the wavelengths of the hydrogen spectrum using integers.

Balmer was devoted to numerology and was interested in things like how many sheep
were in a flock or the number of steps of a Pyramid. He had reconstructed the design of
the Temple given in Chapters 40-43 of the Book of Ezekiel in the Bible. How then, you
may ask, did he come to select the hydrogen spectrum as a problem to solve?

One day, as it happened, Balmer complained to a friend he had "run out of things to do."
The friend replied: "Well, you are interested in numbers, why don't you see what you can
make of this set of numbers that come from the spectrum of hydrogen?" (In 1871
Ångström had measured the wavelengths of the four lines in the visible spectrum of the
hydrogen atom.)

Balmer published his work in two papers, both published in 1885. The first, titled 'Notiz
über die Spektrallinien des Wasserstoff,' is the source of the equation above. He also
gives the value of the constant (3645.6 x 10¯7 mm.) and discusses its significance:

        "One might call this number the fundamental number of hydrogen; and if
        one should succeed in finding the corresponding fundamental numbers for
        other chemical elements as well, then one could speculate that there exist
        between these fundamental numbers and the atomic weights [of the
        substances] in question certain relations, which could be expressed as
        some function."

He goes on to discuss how the constant determined the limiting wavelength of the lines
described by the Balmer Formula:

        "If the formula for n = 2 is correct for all the main lines of the hydrogen
        spectrum, then it implies that towards the utraviolet end these spectral
        lines approach the wavelength 3645.6 in closer and closer sequence, but
        cannot cross this limit; while at the red end [of the spectrum] the C-line
        [today called H] represents the line of longest possible [wavelength].
        Only if in addition lines of higher order existed, would further lines arise
        in the infrared region."

In this second paper, Balmer shows that his formula applies to all 12 of the known lines
in the hydrogen spectrum. Many of the experimentally measured values were very, very
close to Balmer's values, within 0.1 Å or less. There was at least one line, however, that
was about 4 Å off. Balmer expressed doubt about the experimentally measured value,
NOT his formula! He also correctly predicted that no lines longer than the 6562 x 10¯7
mm. line would be discovered in this series and that the lines converge at 3645.6 x 10¯7
mm.

with m = 2, 3, 4, . . . and n = 1, 2, 3, . . . ; but the two constants change in a particular
pattern.


                                               28
By higher order, he means allow n to take on higher values, such as 3, 4, 5, and so on in
this manner:

                                        n        m
                                        3 4, 5, 6, 7, . . .
                                        4 5, 6, 7, 8, . . .
                                        5 6, 7, 8, 9, . . .

There is also this one, but I'm not sure if Balmer discussed it:

                                        n        m
                                        1 2, 3, 4, 5, . . .

Before leaving Balmer, several points:

        1) Balmer's Formula is entirely empirical. By this I mean that it is not
        derived from theory. The equation works, but no one knew why. That is,
        until a certain person.
        2) That certain person was born October 7, 1885 in Copenhagen. His
        name? Niels Henrik David Bohr.
        3) Be careful when you read about Balmer's Formula in other books.
        Often, a form of the formula using frequency rather than wavelength is
        used.

At first Balmer's formula produced nothing but puzzlement, since no theoretical
explanation seemed possible. In 1890 Johannes Robert Rydberg generalized Balmer's
formula and showed that it had a wider applicability. He introduced the concept of the
wave number v, the reciprocal of the wavelength , and wrote his formula as

                                   v = 1/ = R (1/n12 - 1/n22)

where n1 and n2 are integers and R is now known as the Rydberg constant (value =
10973731.534 m¯1). Later many other atomic spectral lines were found to be consistent
with this formula.

For the lines in the hydrogen spectrum (today called the Balmer series), n1 = 2 and n2
takes on the values 3, 4, 5, 6, . . . . If you try the calculations (I don't mind if you do, I can
wait.), remember to do one over the answer, so as to recover the wavelength.



In 1885, Balmer wrote these prophetic words:




                                               29
       "It appeared to me that hydrogen . . . more than any other substance is
       destined to open new paths to the knowledge of the structure of matter and
       its properties. In this respect the numerical relations among the
       wavelengths of the first four hydrogen spectral lines should attract our
       attention particularly."

In 1913, Niels Bohr will announce what is now call the Bohr Model of the Atom. He will
offer the correct mechanism for the lines in the hydrogen spectrum.

                           Balmer Formula Calculations

The Balmer Formula is:




with n = 2 and m = 2, 3, 4, . . . . Balmer used 3645.6 x 10¯7 mm. as the value of the
constant.



The four calculations shown below generate the wavelengths of the four visible lines of
the hydrogen spectrum. Please feel free to carry out each calculation to verify the
answers.




                           Line No. 1: Modern Symbol = H




                           Line No. 2: Modern Symbol = H




                            Line No. 3: Modern Symbol = H




                           Line No. 4: Modern Symbol = H



                                            30
This ends the visible lines of the hydrogen spectrum. Keep in mind that Balmer discused
two points related to the spectrum:

       "If the formula for n = 2 is correct for all the main lines of the hydrogen
       spectrum, then it implies that towards the utraviolet end these spectral
       lines approach the wavelength 3645.6 in closer and closer sequence, but
       cannot cross this limit; while at the red end [of the spectrum] the C-line
       [today called H] represents the line of longest possible [wavelength].
       Only if in addition lines of higher order existed, would further lines arise
       in the infrared region."

By higher order, he means allow n to take on higher values, such as 3, 4, 5, and so on.

With regard to his first point, the next line with n = 2 would take this form:




This line, named H, is in the ultraviolet region of the spectrum.

With regard to his second point no other series of lines, other than the above, was known
to exist. However, with the Balmer formula, production of wavelengths was quite easy
and, as techniques improved, each other series was discovered.



                            The Bohr Model of the Atom



On June 19, 1912, Niels Bohr wrote to his brother Harald:

       "Perhaps I have found out a little about the structure of atoms."

I. Setting the Stage

The Bohr model of the atom deals specifically with the behavior of electrons in the atom.
In constructing his model, Bohr was presented with several problems.

Problem #1: charged electrons moving in an orbit around the nucleus SHOULD radiate
energy due to the acceleration of the electron in its orbit. The frequency of the emitted
radiation should gradually change as the electron lost energy and spiraled into into the
nucleus. Obviously this was not happening, because the spectral lines of a given element
were sharply defined and unchanging.



                                             31
Problem #2: the spectral lines did not show overtones (or harmonics). These are lines
where the frequency is double, triple and so on of the fundamental frequency. The lines
of the spectrum of each element were scattered about with no apparent pattern, other than
the purely empirical formula of Balmer (which dates to 1885). However, no one knew
what Balmer's formula meant.

J.J. Tomson's model was constructed with full knowledge of problem #1 above. What
Thomson did is to extend the positive charge to the same size as the atom (radius = 10¯8
cm.) and allow the electrons to distribute themselves inside. The calculations for his
"plum pudding" model, published in 1904, showed that the model did produce electron
arrangements that were stable.

However, the Thomson model was conclusively destroyed by Rutherford's 1911 nucleus
paper. (In the future -- 1913 and years later -- other discoveries will be made that the
Thomson model fails to account for, but the Rutherford model does. Of course, Thomson,
Rutherford, Bohr, etc. were not aware of these. There were even efforts in 1914 and 1915
to use the Thomson Model, but these efforts went nowhere.)

The nuclear model of Rutherford's was supported by evidence that could not be refuted.
However, if electrons rotated around a nucleus, they would either rip the atom apart or
self-destruct. Bohr's answer to the above problems appeared in print for the world to see
in July 1913. However, as you can see from Bohr's letter to his brother, the journey to the
answer started much earlier.

Bohr wrote out his ideas to date in a memo to Rutherford sometime in June/July 1912.
This memo (with one missing page) still exists and if you were a qualified scholar, you
could go visit it and read it!! Bohr leaves Manchester right after writing the memo and
goes to Copenhagen and is married on August 1, 1912.

The critical part of Bohr's thinking was the making of two assumptions. Bohr himself
described these assumptions on page 7 of his famous paper:

       "(1) That the dynamical equilibrium of the systems in the stationary states
       can be discussed by help of the ordinary mechanics, while the passing of
       the systems between different stationary states cannot be treated on that
       basis.

       (2) That the latter process is followed by the emission of a homogeneous
       radiation, for which the relation between the frequency and the amount of
       energy emitted is the one given by Planck's theory."

Making the assumptions at all was a bit shaky, but in September 1913, at a meeting of the
British Association for the Advancement of Science, Sir James Jeans remarked:

       "The only justification which can be offered for the moment with regard to
       these hypotheses is the very important one that they work in practice."


                                            32
II. Assumption #1: Electrons Move, yet are Stable

Bohr describes "stationary states" (He never uses the word the modern term "orbit.") on
page 5:

       "According to the above considerations, we are led to assume that these
       configurations will correspond to states of the system in which there is no
       radiation of energy states which consequently will be stationary as long as
       the system is not disturbed from outside."

As an electron moves in a "stationary state" it emits no radiation whatsoever. This
violated a branch of science called electrodynamics (having to do with movement of
charged particles and their amount of energy), but the fact is that the atom is stable and
DOES NOT emit radiation in the manner predicted. It is this branch which predicts the
electron will lose energy and crash into the nucleus (this is the problem #1 mentioned at
the top of the file).

In this assumption is some of Bohr's daring nature. While he realized electrodynamics is
useless (second part of his sentence), he proposed to use "mechanics" to describe the
motion of an electron in its orbit (first part of the sentence). Mechanics deals with things
like inertia, momemtum and other features of movement not involving electrical charges.
He was willing to throw out well-supported scientific ideas that didn't work, but was also
willing to keep other ideas that allowed him to make calculations.

The justification for Bohr deciding to assume mechanics held in the atom, but
electrodynamics didn't? The results he got had two features: 1) they concurred with
already known results and 2) offered an explanation for why some results were found and
not others.

III. Assumption #2 - Incorporation of Planck's Constant

The "latter process" in assumption #2 is described at the end of assumption #1 -- "the
passing of the systems between different stationary states"

What Bohr proposed is that the atom will emit (or gain) energy as it moves from one
stationary state to another. However, the amounts of energy will not be any old amount,
but only certain, fixed values. Those values will be the DIFFERENCES in energy
between the stationary states.

Bohr says on p. 7:

       "The second assumption is in obvious contrast to the ordinary ideas of
       electrodynamics but appears to be necessary in order to account for
       experimental facts."


                                             33
The experimental facts refered to are the lines in the spectrum of hydrogen.

What Planck had discovered in 1900 was a fundamental limitation on nature. Energy is
not emitted or absorbed in a continuous manner, but rather in small packets of energy
called quanta. Emission and absorption occured in a DIScontinuous manner. In other
words, an atom moved from one energy state to another state in steps. In the
mathematical description of this process there occured a new constant of nature,
discovered by Planck and named after him.

Planck's constant, symbolized by h, was involved in governing HOW MUCH energy a
given quantum had. The amount of energy was directly dependent on the frequency of
the radiation according to the following equation:

                                          E = hν

This famous equation was first announced by Planck in 1900.

Bohr argued that Planck's constant should be used to help account for the stability of the
atom. This reversed the technique of others who were trying to use atomic models to
determine the physical significance of h. Bohr also realized that, if he was correct, his
theory should produce a constant with the units of length. This constant would
characterize the distance of the electron from the nucleus.

Bohr said on page 2:

       The principal difference between the atom-models proposed by Thomson
       and Rutherford consists in the circumstance the forces acting on the
       electrons in the atom-model of Thomson allow of certain configurations
       and motions of the electrons for which the system is in a stable
       equilibrium; such configurations, however, apparently do not exist for the
       second atom-model. The nature of the difference in question will perhaps
       be most clearly seen by noticing that among the quantities characterizing
       the first atom a quantity appears -- the radius of the positive sphere -- of
       dimensions of a length and of the same order of magnitude as the linear
       extension of the atom, while such a length does not appear among the
       quantities characterizing the second atom, viz. the charges and masses of
       the electrons and the positive nucleus; nor can it be determined solely by
       help of the latter quantities.

       . . . it seems necessary to introduce in the laws in question a quantity
       foreign to the classical electrodynamics, i. e. Planck's constant, or as it
       often is called the elementary quantum of action. By the introduction of
       this quantity the question of the stable configuration of the electrons in the
       atoms is essentially changed as this constant is of such dimensions and
       magnitude that it, together with the mass and charge of the particles, can
       determine a length of the order of magnitude required.


                                            34
What Bohr wass pointing out is that Rutherford's model (with its constants of mass and
charge) cannot produce a unit of length, but with the introduction of h, such a length
constant could be produced. If you write h2 / me2, you get a value with the units of length
and of the proper magnitude. Here are the numbers (modern values):




This length calculation yielded a value which today is named the Bohr radius. Its modern
value is 5.292 x 10¯11 m and is symbolized ao.

                            The Bohr Model of the Atom



Let's stop and review for a moment. What exactly is the Bohr Model of the Atom?

The Bohr model has the following features:

       1) There is a nucleus (this was Rutherford's discovery).
       2) The electrons move about the nucleus in "stationary states" which are
       stable, that is, NOT radiating energy.
       3) When an electron moves from one state to another, the energy lost or
       gained is done so ONLY in very specific amounts of energy, not just any
       old amount.
       4) Each line in a spectrum is produced when an electron moves from one
       stationary state to another.

Today, we call this model an example of a "quantized" atom. The term "quantum" was
introduced by Planck to describe a small bundle of energy. So a quantized atom being
stimulated is shooting out trillions of quanta (plural) of energy per second.

One way to think about the quanta of energy streaming out is to think of a machine gun
shooting out thousands of bullets per second. It SEEMS like a stready, unbroken stream
of metal, but it is not. Each bullet is a "quantum."

Another example is water coming out of a hose. It SEEMS like a steady, unbroken stream
of water, but we know it is just trillions and trillions of tiny, individual water molecules.
The idea is the same with the energy quantum.

Still another example is a stream of gas shooting out a nozzle. It SEEMS like a steady,
unbroken stream of gas, but in reality it is trillions and trillonsof individual gas molecules
all moving in the same direction. Get the idea now??

Last example. In 1905, Albert Einstein wrote an article in 1905 titled "On a Heuristic
Point of View about the Creation and Conversion of Light." In it he uses Planck's idea of


                                             35
a quantum to explain something called the photoelectric effect. He wound up showing
that certain equations governing energy behavior are exactly the same as those for a gas
in the same volume. By analogy, then, energy could be treated the same way as a gas --
as a collection of particles moving around within the volume. He wrote:

       From this we then conclude:

       Monochromatic radiation of low density behaves -- as long as Wein's
       radiation formula is valid -- in a thermodynamic sense, as if it consisted of
       mutually independent energy quanta . . . .

Two paragraphs later, in referring to "monochromatic radiation," he uses the phrase
"discontinuous medium consisting of energy quanta."



Feature #4 above is Bohr's explanation of the mechanism for the production of lines in
the hydrogen spectrum. The story surrounding this is interesting. At least, I think so.

Bohr was married, as I said, in August, 1912. He and his wife did not honeymoon in
Norway as planned, but returned more-or-less immediately to Manchester. The new term
was beginning and Bohr was behind schedule in finishing some work that Rutherford had
assigned to him. Bohr finished that work and eventually the term ended, ending Bohr's
year-long government grant for study.

Bohr even fell behind schedule in writing the paper I am discussing. He wrote a note to
Rutherford on November 4, 1912 apologizing for the time he was taking "to finish my
paper on the atoms and send it to you." By January 1913, Bohr and his wife were back in
Copenhagen where he assumed his new position as assistant to the physics professor.
This slowed down the work on the atomic model paper, but it was not abandonded.

The next event in this story (which is not the complete one, only highlights) is a January
31, 1913 letter to Rutherford in which Bohr excluded the "calculation of frequencies
corresponding to the lines of the visible spectrum." However, the paper Bohr mailed to
Rutherford on March 6, 1913 contained the correct mechanism for the production of the
lines of the spectrum. What happened?

The letter shows Bohr HAD been thinking about spectra. It was probably here (in the first
week of February 1913) that (according to Bohr's recollection in 1954) he was asked by
Hans Marius Hansen (a yound Danish physicist) how Bohr's new atomic model would
explain the hydrogen spectrum. Bohr's reply was that he had not seriously considered the
issue, believing the answer to be impossibly complex. (Remember that by 1913, several
thousand lines, of different elements, were known AND many of these lines exhibited
bizarre splittings - called the Zeeman effect -- in magnetic fields. No one, and I mean no
one, had any answers for what was going on.)




                                            36
Hansen disputed Bohr's position and insisted Bohr look up Balmer's work. Bohr did so
and soon had the answer for how the lines are produced. In the 1954 interview mentioned
above, Bohr said:

       "As soon as I saw Balmer's formula, the whole thing was immediately
       clear to me."

Up to that point, Bohr had not been aware of the Balmer formula. However, it was
quickly incorporated into his paper and the final product was shipped off to Rutherford.

So what is the mechanism that makes the lines in the spectrum? On page 11, after
discussing the mechanism from p. 8 onward, Bohr said:

       " . . . that the lines correspond to a radiation emitted during the passing of
       the system between two different stationary states."

Later, on p. 14, he wrote:

       "We are thus led to assume that the interpretation of the equation (2) is not
       that the different stationary states correspond to an emission of different
       numbers of energy-quanta, but that the frequency of the energy emitted
       during the passing of the system from a state in which no energy is yet
       radiated out to one of the different stationary states, . . . ."

So, in other words, each line is equal to the DIFFERENCE in energy as an electron
moves between two stationary states. And, if you read Bohr's paper, at the bottom of page
8, you will see two equations where he writes Wτ2 minus Wτ1. This subtraction, of course,
yields an answers which we call a "difference."

This is also the reason why the harmonics (see back to problem #2) were not observed.
Harmonics are observed when an object (such as an electron) resonates at a specific
frequency, called the fundamental. Overtones of twice the frequency, three times the
frequency and so on are also observed. Here the frequency of the line in the spectrum had
to do with energy gained or lost as the electron moved from one stationary state to
another.The line is an energy DIFFERENCE!! The fundamental frequency of the electron
was not involved at all with the lines!! No one until Bohr, not even Planck or Einstein,
had thought to challenge the idea that spectral lines were produced by electrons emitting
the specific frequenceies of each line.

By the way, Bohr never discusses exactly how an electron moves between two stationary
states. How does the electron know to go to another stationary state of a specific energy
amount? Why that amount and not some other amount? How does the electron know to
stop at the right energy amount? Maybe Bo knows, but Bohr certainly didn't.

Bohr sent a second draft to Rutherford about two weeks after the first since he had "found
it necessary to introduce some small alterations and additions." He then traveled to


                                             37
Manchester to persuade Rutherford that the paper would harmed greatly by any reduction
in length. On May 10, Bohr returned the final, corrected proof to Rutherford and this
became the published version.

In August 1914 (slightly more that a year after Bohr's paper was published), Rutherford
said:

       "N. Bohr has faced the difficulties [of atomic structure] by bringing in the
       idea of the quantum. At all events there is something going on which is
       inexplicable by the older mechanics."

Of course, Rutherford was completely correct. The "older mechanics" had to give way to
the "quantum mechanics." Bohr would lead a revolution in thinking during the 1920s and
1930s that continues to be profitably mined even today.

At age eighty, J.J. Thomson, stooped by age, but still sharp of mind, wrote:

       "At the end of 1913 Niels Bohr published the first of a series of researches
       on spectra, which it is not too much to say have in some departments of
       spectroscopy changed chaos into order, and which were, I think, the most
       valuable contributions which quantum theory has ever made to physical
       science."




                             Quantum Numbers
There are four quantum numbers; their symbols are n, l, m and s. EVERY electron in an
atom has a specific, unique set of these four quantum numbers. The story behind how
these numbers came to be discovered is a complex one indeed and is one best left for
another day.

A warning before I proceed: this is a 100% non-mathematical discussion. The equations
governing electron behavior in atoms are complex. This area of study generated LOTS of
Nobel Prizes and the reasoning leading to the above mentioned equations is sophisticated
and sometimes quite subtle.

Just keep this in mind: EVERY electron's behavior in an atom is governed by a set of
equations and that n, l, m, and s are values in those equations. EVERY electron in an
atom has a unique set of quantum numbers.

Lastly, I'm going to reserve to another discussion what these numbers mean. I will just
describe their existence and the rules for how to determine them in this tutorial. The next
tutorial will start with hydrogen and assign quantum numbers to its electron, then proceed
to helium and do the same, then lithium, beryllium, and so on.



                                            38
I. The Principal Quantum Number (signified by the letter 'n'): This quantum number
was the first one discovered and it was done so by Niels Bohr in 1913. Bohr thought that
each electron was in its own unique energy level, which he called a "stationary state," and
that each electron would have a unique value of 'n.'

In this idea, Bohr was wrong. It very quickly was discovered that more than one electron
could have a given 'n' value. For example, it was eventually discovered that when n=3,
eighteen different electrons could have that value.

Keep in mind that it is the set of four quantum numbers that is important. As you will see,
each of the 18 electrons just mention will have its own unique set of n, l, m, and s.

Finally, there is a rule for what values 'n' can assume. It is:

                                    n = 1, 2, 3, and so on.

n will always be a whole number and NEVER less than one.

One point: n does not refer to any particular location in space or any particular shape. It is
one component (of four) that will uniquely identify each electron in an atom.

II. The Azimuthal Quantum Number (signified by the letter 'l'): about 1914-1915,
Arnold Sommerfeld realized that Bohr's 'n' was insufficient. In other words, more
equations were needed to properly describe how electrons behaved. In fact, Sommerfeld
realized that TWO more quantum numbers were needed.

The first of these is the quantum number signified by 'l.' When Sommerfeld started this
work, he used n' (n prime), but he shifted it to 'l' after some years. I'm not sure why, but it
seems easier to print l than n prime and what if the printer (of a textbook) accidently
dumps a few prime symbols, leaving just the letter 'n?' Ooops!

The rule for selecting the proper values of 'l' is as follows:

                                      l = 0, 1, 2, . . . , n-1

l will always be a whole number and will NEVER be as large as the 'n' value it is
associated with.

III. The Magnetic Quantum Number (signified by the letter 'm'): this quantum
number was also discovered by Sommerfeld in the same 1914-1915 time frame. I don't
think he discovered one and then the other, I think that him realizing the need for two
runs together somewhat. I could be wrong in this, so don't take my word for it!

The rule for selecting m is as follows:

    m starts at negative 'l,' runs by whole numbers to zero and then goes to positive 'l.'


                                                39
For example, when l = 2, the m values used are -2, -1, 0, +1, +2, for a total of five values.

IV. The Spin Quantum Number (signified by the letter 's'): spin is a property of
electrons that is not related to a sphere spinning. It was first thought to be this way, hence
the name spin, but it was soon realized that electrons cannot spin on their axis like the
Earth does on its axis. If the electron did this, its surface would be moving at about ten
times the speed of light (if memory serves correctly!). In any event, the electron's surface
would have to move faster than the speed of light and this isn't possible.

In 1925, Wolfgang Pauli demonstrated the need for a fourth quantum number. He closed
the abstract to his paper this way:

       "On the basis of these results one is also led to a general classification of
       every electron in the atom by the principal quantum number n and two
       auxiliary quantum numbers k1 and k2 to which is added a further quantum
       number m1 in the presence of an external field. In conjunction with a
       recent paper by E. C. Stoner this classification leads to a general quantum
       theoretical formulation of the completion of electron groups in atoms."

In late 1925, two young researchers named George Uhlenbeck and Samuel Goudsmit
discovered the property of the electron responsible for the fourth quantum number being
needed and named this property spin.

The rule for selecting s is as follows:

    after the n, l and m to be used have been determined, assign the value +1/2 to one
  electron, then assign -1/2 to the next electron, while using the same n, l and m values.

For example, when n, l, m = 1, 0, 0; the first s value used is +1/2, however a second
electron can also have n, l, m = 1, 0, 0; so assign -1/2 to it.

                        Quantum Numbers H to Ne
There are four quantum numbers: n, l, m, and s. Each one is a particular factor in an
equation describing a property of the electron. At this introductory level, the equations
are not needed. The value of each quantum number is assigned to each electron in an
atom by a "building up" process. Niels Bohr called this process the "Aufbau" principle:
aufbau means "building up."

n is ALWAYS the starting point for building up a series of quantum numbers. Each
quantum number is then assigned according to a set of rules, each of which took years of
study to finally determine. The rules ARE NOT just any old arbitrary ones; they have
been determined from a study of nature. Remember the rules:

       (1) n = 1, 2, 3, and so on.
       (2) l = 0, 1, 2, . . . , n-1


                                             40
       (3) m starts at negative 'l,' runs by whole numbers to zero and then goes to
       positive 'l.'
       (4) after the n, l and m to be used have been determined, assign the value
       +1/2 to one electron, then assign -1/2 to the next electron, while using the
       same n, l and m values.

Also, keep in mind that we use only one n, l, m, and s value each to make a set of four
quantum numbers for each electron. It is the set that uniquely identifies each electron.

Last point: the last column in each table below is called "Orbital Name." As you are
reading this tutorial, you may not yet know what an orbital is. That's OK, but please
understand the concept called "orbital" is an important one. Here's a real simple
description that ignores lots of details: each orbital is a region of space around the
nucleus which contains a MAXIMUM of two electrons. Realize that it's more complex
than that, but the above description is good enough for now. I hope!!



Hydrogen - one electron

First Electron

n=1
l=0
m=0

In each case, note that we start with the smallest value of n, l, or m possible. Make sure
you look over the rules to see how each value was arrived at. l starts at zero and goes to
n-1, which is zero since we get 1-1 = 0, when using n = 1. When l = 0, there is only one
possible choice for m, which must be zero.

s = +1/2

This completes the four quantum numbers for the single electron possessed by hydrogen.
I shall build up a table like this:

                   Atomic                                         Orbital
                              Element     n        l   m    s
                   Number                                         Name
                      1      Hydrogen     1        0   0   +1/2     1s

Helium - two electrons

First Electron




                                              41
n=1
l=0
m=0
s = +1/2

The first electron in helium has exactly the same four quantum number of the first
electron in hydrogen. However, helium has TWO electrons. So we "build up" from the
previous electrons by adding one more.

Second Electron

n=1
l=0
m=0
s = -1/2

Notice the same n, l, and m values, but s has shifted from positive 1/2 to negative 1/2.
This was the problem Pauli saw in 1925. Three quantum numbers was insufficient to
UNIQUELY identify each electron, but a fourth one (the one called s) did the trick.

                   Atomic                                         Orbital
                              Element     n        l   m    s
                   Number                                         Name
                       2        Helium    1        0   0   +1/2
                                                                    1s
                                          1        0   0   -1/2

Lithium - three electrons

The first two electrons quantum numbers' are EXACTLY the same as the two in helium:

                               1, 0, 0, +1/2 and 1, 0, 0, -1/2

Third Electron: here's where we "build up" by adding one more electron.

However, we are now presented with a problem. All the values with n = 1 have been used
up, but we have only accounted for two of lithium's three electrons. What to do about the
third?

Answer: start with the NEXT n value; n = 2. However, there is a problem with l; do we
use l = 0 or l =1, since both are possible with n = 2?

Answer: start with the lowest value first, so that means using l = 0. (Don't worry, we will
use l = 1 soon enough.)

Figuring out m should be easy; when l = 0, m can only equal 0. So n, l, m for the third
electron is 2, 0, 0. I'll add in s in the table below.


                                              42
                    Atomic                                         Orbital
                               Element     n        l   m    s
                    Number                                         Name
                       3        Lithium    1        0   0   +1/2
                                                                     1s
                                           1        0   0   -1/2
                                           2        0   0   +1/2     2s

Beryllium - four electrons

In the building up process, we go one electron at a time. Therefore, we will use the three
from lithium and add one more.

Fourth Electron

n=2
l=0
m=0
s = -1/2

Notice the same n, l, and m values as the third electron, but s has shifted from positive
1/2 to negative 1/2.

                   Atomic                                          Orbital
                               Element     n        l   m    s
                   Number                                          Name
                       4      Beryllium    1        0   0   +1/2
                                                                      1s
                                           1        0   0   -1/2
                                           2        0   0   +1/2
                                                                      2s
                                           2        0   0   -1/2


Pretty easy, heh? It stays easy, if you follow the rules. With beryllium, we have
exhausted the possibilities for the n = 2; l = 0 combination. However, when n = 2, l can
take on another value, namely l = 1. This has consequences for the m value as well and,
after we finish, there will be six electrons that have a combination of n = 2 and l = 1.

Here's the rule for m again: starts at negative 'l,' runs by whole numbers to zero and then
goes to positive 'l.' Since l = 1, we start with -1, go to zero and end up at +1. This gives us
three values for m when l = 1. Hopefully you can see that, since s takes on +1/2 and -1/2,
we will wind up with six sets of quantum numbers.

Warning: there's going to be a new rule introduced after boron. So prepare yourself
because, just as you thought it was getting easy, there gets added some new stuff. By the
way, us mean old teachers didn't make all this stuff up to torture poor chemistry students.
Nature really does do what I will explain below. Here's boron:


                                               43
Boron - five electrons

Following the usual pattern, I've repeated the previous four electrons. As we go on to the
l = 1 values, keep in mind that we will start with the lowest value of m, namely negative
one.

                   Atomic                                          Orbital
                              Element      n        l   m    s
                   Number                                          Name
                       5         Boron     1        0   0   +1/2
                                                                     1s
                                           1        0   0   -1/2
                                           2        0   0   +1/2
                                                                     2s
                                           2        0   0   -1/2
                                           2        1   -1 +1/2     2px

Eventually, I will wind up with three orbital names. 2px is just the first, x meaning the x-
axis. Next will be 2py, for the y-axis and the last name used will be 2pz, for the z-axis.
These three orbitals are oriented at 90° to each other.



Hund's Rule (named for Fredrich Hund) is the name of the new rule. This rule concerns
the relationship between the l and m quantum numbers. When l = 0, m can only equal
zero and Hund's Rule does not show up. However, now that we have reached l = 1, m can
take on multiple values. Hund's Rule concerns the order in which we assign the l and m
values.

By the way, I'm going to avoid a technical statement of Hund's Rule for the moment. I'll
discuss how it works first.

Hund's Rule means that we will use each possible l, m combination ONCE before going
back and using it a second time. Here are the three possible l, m combos when l = 1:

                                            l m
                                           1 -1
                                           1 0
                                           1 +1

For boron, we have used the l, m combination of 1, -1. The key is to see that Hund's Rule
requires we go on to the NEXT l, m combination for the next element: carbon.



                                               44
Carbon - six electrons

Following the usual pattern, I've repeated the previous five electrons. As we continue on
with the l = 1 values, keep in mind that Hund's Rule will affect how we assign the next m
value.

                   Atomic                                          Orbital
                               Element     n        l   m    s
                   Number                                          Name
                       6        Carbon     1        0   0   +1/2
                                                                     1s
                                           1        0   0   -1/2
                                           2        0   0   +1/2
                                                                     2s
                                           2        0   0   -1/2
                                           2        1   -1 +1/2     2px
                                           2        1   0   +1/2    2py

Nitrogen - seven electrons

Since we still have not first used all possible l, m values ONCE, we go on to the next l, m
combination.

                   Atomic                                          Orbital
                               Element     n        l   m    s
                   Number                                          Name
                       7       Nitrogen    1        0   0   +1/2
                                                                     1s
                                           1        0   0   -1/2
                                           2        0   0   +1/2
                                                                     2s
                                           2        0   0   -1/2
                                           2        1   -1 +1/2     2px
                                           2        1   0   +1/2    2py
                                           2        1   +1 +1/2     2pz

2px, 2py and 2pz are three different orbitals, each one capable of holding two electrons.
Notice how, in nitrogen, each of the three orbitals is filled up HALF-WAY (that is, with
one electron) before we go back and fill up each orbital with the second electron.

This "half-filled orbital" has definite chemical consequences. Remember it well. Also,
using 2px first, then going to y and then z is purely convention. The x, y, z order is not of
consequence in the above examples. However keep in mind the using each letter ONCE
first being using it for the second electron is important.

Oxygen - eight electrons



                                               45
Now that we have used each l, m combination once, we proceed to go back and use each
combo the second time. For oxygen to neon, I've marked which electron is the one added.

                  Atomic                                         Orbital
                             Element    n        l   m     s
                  Number                                         Name
                       8      Oxygen    1        0   0    +1/2
                                                                   1s
                                        1        0   0    -1/2
                                        2        0   0    +1/2
                                                                   2s
                                        2        0   0    -1/2
                                        2        1   -1 +1/2
                  this one                                        2px
                                 --->   2        1   -1   -1/2
                   added
                                        2        1   0    +1/2    2py
                                        2        1   +1 +1/2      2pz

Fluorine - nine electrons

                  Atomic                                         Orbital
                             Element    n        l   m     s
                  Number                                         Name
                       9     Fluorine   1        0   0    +1/2
                                                                   1s
                                        1        0   0    -1/2
                                        2        0   0    +1/2
                                                                   2s
                                        2        0   0    -1/2
                                        2        1   -1 +1/2
                                                                  2px
                                        2        1   -1   -1/2
                                        2        1   0    +1/2
                  this one                                        2py
                                 --->   2        1   0    -1/2
                   added
                                        2        1   +1 +1/2      2pz

Neon - ten electrons

                  Atomic                                         Orbital
                             Element    n        l   m     s
                  Number                                         Name
                       10       Neon    1        0   0    +1/2
                                                                   1s
                                        1        0   0    -1/2
                                        2        0   0    +1/2
                                                                   2s
                                        2        0   0    -1/2


                                            46
                                         2        1   -1 +1/2
                                                                  2px
                                         2        1   -1   -1/2
                                         2        1   0    +1/2
                                                                  2py
                                         2        1   0    -1/2
                                         2        1   +1 +1/2
                   this one                                       2pz
                                  --->   2        1   +1 -1/2
                    added

We have now completed all possible values for n = 1 AND n = 2. Starting with element
11, sodium, we will proceed on to n = 3. When we finish, we will have used l = 0, l = 1
(and applied Hund's Rule again) and then, before going on to l = 3, we will hit another
interesting twist that nature has handed us. We will wind up going on to n = 4 and then
coming back to finish n = 3. It will be fun!




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